I would like to verify that
$$\int_{\mathrm{SU}(2)}\mathrm{d}g\,\delta(g)=1$$
where the "delta-function" is defined via
$$\delta(g):=\sum_{j\in\mathbb{N}_{0}/2}(1+2j)\chi^{j}(g)$$ where $\chi^{j}$ are the characters of the irreducible unitary spin-j representation of $\mathrm{SU}(2)$ with dimension $2j+1$.
My attempt: The normalized Haar measure of SU(2) can be written in terms of Euler angles $\alpha,\beta,\gamma$ as
$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,f(g)=\frac{1}{8\pi^{2}}\int_{0}^{2\pi}\,\mathrm{d}\alpha\,\int_{0}^{\pi}\,\mathrm{d}\beta\,\mathrm{sin}(\beta)\,\int_{0}^{2\pi}\,\mathrm{d}\gamma\,f(\alpha,\beta,\gamma).$$
Furthermore, the characters $\chi_{j}$ can be written as
$$\chi^{j}(g(\alpha,\beta,\gamma))=\frac{\mathrm{sin}((2j+1)\beta/2)}{\mathrm{sin}(\beta/2)}$$
according to this wikipedia article. Hence, we have to show that
$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g)=\int_{0}^{\pi}\,\mathrm{d}\beta\,\mathrm{sin}(\beta)\,\bigg (\sum_{j=0}^{\infty}\frac{j+1}{2}\frac{\mathrm{sin}((j+1)\beta/2)}{\mathrm{sin}(\beta/2)}\bigg )\overset{!}{=}1$$
(Here, I also rescaled the sum 2j$\to$ j). Now, since I am a physicist, I am allowed to change the infinite sum with the integral (maybe the error lies there; maybe I should check with dominated convergence theorem). Let us look at the integrals appearing in the expression above:
$$\int_{0}^{\pi}\,\mathrm{d}\beta\,\mathrm{sin}(\beta)\,\frac{\mathrm{sin}((j+1)\beta/2)}{\mathrm{sin}(\beta/2)}=\begin{cases}2 &\text{if j=0}\\ 4\frac{1+j-\mathrm{cos}(j\pi/2)}{j(2+j)} &\text{else}\end{cases}$$
Hence, I get something like
$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g)=1+\sum_{j=1}^{\infty}\frac{j+1}{2}\frac{4(1+j-\mathrm{cos}(j\pi/2))}{j(2+j)}$$
However, the series on the right-hand side is clearly not convergent: A quick mathematica calculation shows
$$\sum_{j=1}^{N}\frac{j+1}{2}\frac{4(1+j-\mathrm{cos}(j\pi/2))}{j(2+j)}=\begin{cases}\frac{241}{11}\cong 22 &\text{N=10}\\\frac{1040350}{5151}\cong 202 &\text{N=100}\\\frac{1004003500}{501501}\cong 2002 &\text{N=1000}\\\vdots\end{cases}$$
I think the error lies indeed in exchanging the series with the integral. If I look at the series in
$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,\delta(g)=\int_{0}^{\pi}\,\mathrm{d}\beta\,\mathrm{sin}(\beta)\,\bigg (\sum_{j=0}^{\infty}\frac{j+1}{2}\frac{\mathrm{sin}((j+1)\beta/2)}{\mathrm{sin}(\beta/2)}\bigg )$$
we have that
$$\sum_{j=0}^{\infty}(j+1)\mathrm{sin}((j+1)\beta/2)=\sum_{j=1}^{\infty}j\cdot \mathrm{sin}\bigg(j\cdot\frac{\beta}{2}\bigg )$$
The series on the right-hand side seems to be zero, when evaluated with Mathematica, but I can't see why this is the case...